For the decomposition of a carbonate, $Δ H^{\circ} = + 178.0 kJ mo l^{- 1}$ and $Δ S^{\circ} = + 165 J mo l^{- 1} K^{- 1}$ , both assumed independent of temperature. At what temperature (in K, nearest integer) does the equilibrium constant become equal to 1?

Numerical answer type, answer in K, tolerance 1.

Answer: 1079 K

K = 1

means

Δ G^{\circ} = 0 = Δ H^{\circ} - T Δ S^{\circ}

T = Δ H^{\circ} /Δ S^{\circ} = 178000/165 = 1078.79 K

, i.e. 1079 K. Above this temperature

K > 1